Determine the sum $sum_n=1^infty left( frac1(4n)^2-1-frac1(4n+2)^2+1right)$ using the Fourier series of $|cos x|$How much makes $sumlimits_i=-infty^infty frac1i2pi+x$?Fourier series $sum_m=0^infty fraccos (2m+1)x2m+1$Calculate the sum using Fourier series of $big|cos(fracx2)big|$.Find the sum of a series using Fourier seriesFourier series question - represent $x$ as a series of $cos$Does the sum $sum_n=1^inftya_nb_n$ converge(fourier series coefficients)?A proof of a known identity using Fourier seriesDeriving sum $sum_n=1^+infty frac(-1)^n36n^2-25$ from Fourier seriesCan you obtain the Fourier Series of $cos x$ from the Fourier Series of $cos(x/2)$?calculating sum of a series using fourier seriesComputing $sum_n = 1^+ infty frac1n^2 + 1$ using Fourier series.
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Determine the sum $sum_n=1^infty left( frac1(4n)^2-1-frac1(4n+2)^2+1right)$ using the Fourier series of $|cos x|$
How much makes $sumlimits_i=-infty^infty frac1i2pi+x$?Fourier series $sum_m=0^infty fraccos (2m+1)x2m+1$Calculate the sum using Fourier series of $big|cos(fracx2)big|$.Find the sum of a series using Fourier seriesFourier series question - represent $x$ as a series of $cos$Does the sum $sum_n=1^inftya_nb_n$ converge(fourier series coefficients)?A proof of a known identity using Fourier seriesDeriving sum $sum_n=1^+infty frac(-1)^n36n^2-25$ from Fourier seriesCan you obtain the Fourier Series of $cos x$ from the Fourier Series of $cos(x/2)$?calculating sum of a series using fourier seriesComputing $sum_n = 1^+ infty frac1n^2 + 1$ using Fourier series.
$begingroup$
I need to determine the sum $$sum_n=1^infty left( frac1(4n)^2-1-frac1(4n+2)^2+1right)$$ using the Fourier series of $lvert cos xrvert$ on the interval $ [-pi,pi]$.
I have already calculated Fourier series and I get this:
$$lvert cos xrvert= frac2pi+sum_n=2^inftyfrac4pifraccos(nfracpi2)1-n^2cos(nx)$$
I do not know how to manipulate the Fourier series to get that specific sum.
I tried this but did not get me anywhere.
$$sum_n=2^inftyfrac4pifracfrac12cos(nfracpi2-nx)-frac12cos(nfracpi2+nx)1-n^2$$
$$-sum_n=2^inftyfrac4pifraccos(nfracpi2-nx)2n^2-2-fraccos(nfracpi2+nx)2n^2-2$$
real-analysis sequences-and-series fourier-series
asked Mar 21 at 10:40
Davor Davor
112
$endgroup$
|
show 2 more comments
$begingroup$
I need to determine the sum $$sum_n=1^infty left( frac1(4n)^2-1-frac1(4n+2)^2+1right)$$ using the Fourier series of $lvert cos xrvert$ on the interval $ [-pi,pi]$.
I have already calculated Fourier series and I get this:
$$lvert cos xrvert= frac2pi+sum_n=2^inftyfrac4pifraccos(nfracpi2)1-n^2cos(nx)$$
I do not know how to manipulate the Fourier series to get that specific sum.
I tried this but did not get me anywhere.
$$sum_n=2^inftyfrac4pifracfrac12cos(nfracpi2-nx)-frac12cos(nfracpi2+nx)1-n^2$$
$$-sum_n=2^inftyfrac4pifraccos(nfracpi2-nx)2n^2-2-fraccos(nfracpi2+nx)2n^2-2$$
real-analysis sequences-and-series fourier-series
asked Mar 21 at 10:40
Davor Davor
112
$endgroup$
$begingroup$
$cos(n fracpi2) = 1 forall n in mathbbN$
$endgroup$
– Max
Mar 21 at 11:01
2
$begingroup$
@Max This is by factor 4 far from truth.
$endgroup$
– user
Mar 21 at 11:04
$begingroup$
Sorry, lack of coffee. Was meant to be $=0$.
$endgroup$
– Max
Mar 21 at 11:13
$begingroup$
Its not 0, it can be 0, -1 or 1 depending on n
$endgroup$
– Davor
Mar 21 at 11:20
$begingroup$
Oh dammit, i'm getting coffee now! Sorry for the nonsense.
$endgroup$
– Max
Mar 21 at 11:30
|
show 2 more comments
$begingroup$
I need to determine the sum $$sum_n=1^infty left( frac1(4n)^2-1-frac1(4n+2)^2+1right)$$ using the Fourier series of $lvert cos xrvert$ on the interval $ [-pi,pi]$.
I have already calculated Fourier series and I get this:
$$lvert cos xrvert= frac2pi+sum_n=2^inftyfrac4pifraccos(nfracpi2)1-n^2cos(nx)$$
I do not know how to manipulate the Fourier series to get that specific sum.
I tried this but did not get me anywhere.
$$sum_n=2^inftyfrac4pifracfrac12cos(nfracpi2-nx)-frac12cos(nfracpi2+nx)1-n^2$$
$$-sum_n=2^inftyfrac4pifraccos(nfracpi2-nx)2n^2-2-fraccos(nfracpi2+nx)2n^2-2$$
real-analysis sequences-and-series fourier-series
asked Mar 21 at 10:40
Davor Davor
112
$endgroup$
I need to determine the sum $$sum_n=1^infty left( frac1(4n)^2-1-frac1(4n+2)^2+1right)$$ using the Fourier series of $lvert cos xrvert$ on the interval $ [-pi,pi]$.
I have already calculated Fourier series and I get this:
$$lvert cos xrvert= frac2pi+sum_n=2^inftyfrac4pifraccos(nfracpi2)1-n^2cos(nx)$$
I do not know how to manipulate the Fourier series to get that specific sum.
I tried this but did not get me anywhere.
$$sum_n=2^inftyfrac4pifracfrac12cos(nfracpi2-nx)-frac12cos(nfracpi2+nx)1-n^2$$
$$-sum_n=2^inftyfrac4pifraccos(nfracpi2-nx)2n^2-2-fraccos(nfracpi2+nx)2n^2-2$$
real-analysis sequences-and-series fourier-series
real-analysis sequences-and-series fourier-series
asked Mar 21 at 10:40
Davor Davor
112
asked Mar 21 at 10:40
Davor Davor
112
edited Mar 22 at 9:31
Davor
asked Mar 21 at 10:40
Davor Davor
112
asked Mar 21 at 10:40
Davor Davor
112
asked Mar 21 at 10:40
Davor Davor
112
112
$begingroup$
$cos(n fracpi2) = 1 forall n in mathbbN$
$endgroup$
– Max
Mar 21 at 11:01
2
$begingroup$
@Max This is by factor 4 far from truth.
$endgroup$
– user
Mar 21 at 11:04
$begingroup$
Sorry, lack of coffee. Was meant to be $=0$.
$endgroup$
– Max
Mar 21 at 11:13
$begingroup$
Its not 0, it can be 0, -1 or 1 depending on n
$endgroup$
– Davor
Mar 21 at 11:20
$begingroup$
Oh dammit, i'm getting coffee now! Sorry for the nonsense.
$endgroup$
– Max
Mar 21 at 11:30
|
show 2 more comments
$begingroup$
$cos(n fracpi2) = 1 forall n in mathbbN$
$endgroup$
– Max
Mar 21 at 11:01
2
$begingroup$
@Max This is by factor 4 far from truth.
$endgroup$
– user
Mar 21 at 11:04
$begingroup$
Sorry, lack of coffee. Was meant to be $=0$.
$endgroup$
– Max
Mar 21 at 11:13
$begingroup$
Its not 0, it can be 0, -1 or 1 depending on n
$endgroup$
– Davor
Mar 21 at 11:20
$begingroup$
Oh dammit, i'm getting coffee now! Sorry for the nonsense.
$endgroup$
– Max
Mar 21 at 11:30
$begingroup$
$cos(n fracpi2) = 1 forall n in mathbbN$
$endgroup$
– Max
Mar 21 at 11:01
$begingroup$
$cos(n fracpi2) = 1 forall n in mathbbN$
$endgroup$
– Max
Mar 21 at 11:01
2
2
$begingroup$
@Max This is by factor 4 far from truth.
$endgroup$
– user
Mar 21 at 11:04
$begingroup$
@Max This is by factor 4 far from truth.
$endgroup$
– user
Mar 21 at 11:04
$begingroup$
Sorry, lack of coffee. Was meant to be $=0$.
$endgroup$
– Max
Mar 21 at 11:13
$begingroup$
Sorry, lack of coffee. Was meant to be $=0$.
$endgroup$
– Max
Mar 21 at 11:13
$begingroup$
Its not 0, it can be 0, -1 or 1 depending on n
$endgroup$
– Davor
Mar 21 at 11:20
$begingroup$
Its not 0, it can be 0, -1 or 1 depending on n
$endgroup$
– Davor
Mar 21 at 11:20
$begingroup$
Oh dammit, i'm getting coffee now! Sorry for the nonsense.
$endgroup$
– Max
Mar 21 at 11:30
$begingroup$
Oh dammit, i'm getting coffee now! Sorry for the nonsense.
$endgroup$
– Max
Mar 21 at 11:30
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
The sum of the first term is
$$
beginalign
sum_n=1^inftyfrac1(4n)^2-1
&=frac12sum_n=1^inftyleft(frac14n-1-frac14n+1right)\
&=frac12left[1+sum_ninmathbbZfrac14n-1right]\
&=frac12+frac18sum_ninmathbbZfrac1n-frac14\
&=frac12-fracpi8cotleft(fracpi4right)\[6pt]
&=frac12-fracpi8tag1
endalign
$$
The sum of the second term is
$$
beginalign
sum_n=1^inftyfrac1(4n+2)^2+1
&=frac i2sum_n=1^inftyleft(frac1i-(4n+2)+frac1i+(4n+2)right)\
&=frac i2left[sum_ninmathbbZfrac1i+(4n+2)-frac1i+2-frac1i-2right]\
&=frac i8sum_ninmathbbZfrac1n+frac2+i4-frac15\
&=fracpi i8cotleft(pifrac2+i4right)-frac15\[3pt]
&=fracpi8tanhleft(fracpi4right)-frac15tag2
endalign
$$
where we have used $(7)$ from this answer (which uses complex analysis) or Theorem $2$ from this answer (which uses Fourier Series).
Subtracting $(2)$ from $(1)$ yields
$$
sum_n=1^inftyleft(frac1(4n)^2-1-frac1(4n+2)^2+1right)=frac710-fracpi8-fracpi8tanhleft(fracpi4right)tag3
$$
answered Mar 21 at 20:00
robjohn♦robjohn
270k27313641
$endgroup$
$begingroup$
Mathematica agrees with this answer.
$endgroup$
– robjohn♦
Mar 21 at 20:01
$begingroup$
Thank you sir, I have checked your answer, it is correct, just in the first sum is (4n)^2 and not 4n^2. Also my answer is also correct, i have checked it with wolfram mathematica also on yt youtube.com/watch?v=bG6bpYlQx2s
$endgroup$
– Davor
Mar 22 at 9:13
$begingroup$
@Davor: you mean that the first sum should be $sumlimits_n=1^inftyfrac116n^2-1$? Are you going to alter the question, or ask another with the correct sum?
$endgroup$
– robjohn♦
Mar 22 at 9:23
$begingroup$
Yes, i have corrected the equation, but it does not matter that much, i was more interested how to do it. We did not learn that approach with complex analysis, only usinf fourier transform
$endgroup$
– Davor
Mar 22 at 9:35
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The sum of the first term is
$$
beginalign
sum_n=1^inftyfrac1(4n)^2-1
&=frac12sum_n=1^inftyleft(frac14n-1-frac14n+1right)\
&=frac12left[1+sum_ninmathbbZfrac14n-1right]\
&=frac12+frac18sum_ninmathbbZfrac1n-frac14\
&=frac12-fracpi8cotleft(fracpi4right)\[6pt]
&=frac12-fracpi8tag1
endalign
$$
The sum of the second term is
$$
beginalign
sum_n=1^inftyfrac1(4n+2)^2+1
&=frac i2sum_n=1^inftyleft(frac1i-(4n+2)+frac1i+(4n+2)right)\
&=frac i2left[sum_ninmathbbZfrac1i+(4n+2)-frac1i+2-frac1i-2right]\
&=frac i8sum_ninmathbbZfrac1n+frac2+i4-frac15\
&=fracpi i8cotleft(pifrac2+i4right)-frac15\[3pt]
&=fracpi8tanhleft(fracpi4right)-frac15tag2
endalign
$$
where we have used $(7)$ from this answer (which uses complex analysis) or Theorem $2$ from this answer (which uses Fourier Series).
Subtracting $(2)$ from $(1)$ yields
$$
sum_n=1^inftyleft(frac1(4n)^2-1-frac1(4n+2)^2+1right)=frac710-fracpi8-fracpi8tanhleft(fracpi4right)tag3
$$
answered Mar 21 at 20:00
robjohn♦robjohn
270k27313641
$endgroup$
$begingroup$
Mathematica agrees with this answer.
$endgroup$
– robjohn♦
Mar 21 at 20:01
$begingroup$
Thank you sir, I have checked your answer, it is correct, just in the first sum is (4n)^2 and not 4n^2. Also my answer is also correct, i have checked it with wolfram mathematica also on yt youtube.com/watch?v=bG6bpYlQx2s
$endgroup$
– Davor
Mar 22 at 9:13
$begingroup$
@Davor: you mean that the first sum should be $sumlimits_n=1^inftyfrac116n^2-1$? Are you going to alter the question, or ask another with the correct sum?
$endgroup$
– robjohn♦
Mar 22 at 9:23
$begingroup$
Yes, i have corrected the equation, but it does not matter that much, i was more interested how to do it. We did not learn that approach with complex analysis, only usinf fourier transform
$endgroup$
– Davor
Mar 22 at 9:35
add a comment |
$begingroup$
The sum of the first term is
$$
beginalign
sum_n=1^inftyfrac1(4n)^2-1
&=frac12sum_n=1^inftyleft(frac14n-1-frac14n+1right)\
&=frac12left[1+sum_ninmathbbZfrac14n-1right]\
&=frac12+frac18sum_ninmathbbZfrac1n-frac14\
&=frac12-fracpi8cotleft(fracpi4right)\[6pt]
&=frac12-fracpi8tag1
endalign
$$
The sum of the second term is
$$
beginalign
sum_n=1^inftyfrac1(4n+2)^2+1
&=frac i2sum_n=1^inftyleft(frac1i-(4n+2)+frac1i+(4n+2)right)\
&=frac i2left[sum_ninmathbbZfrac1i+(4n+2)-frac1i+2-frac1i-2right]\
&=frac i8sum_ninmathbbZfrac1n+frac2+i4-frac15\
&=fracpi i8cotleft(pifrac2+i4right)-frac15\[3pt]
&=fracpi8tanhleft(fracpi4right)-frac15tag2
endalign
$$
where we have used $(7)$ from this answer (which uses complex analysis) or Theorem $2$ from this answer (which uses Fourier Series).
Subtracting $(2)$ from $(1)$ yields
$$
sum_n=1^inftyleft(frac1(4n)^2-1-frac1(4n+2)^2+1right)=frac710-fracpi8-fracpi8tanhleft(fracpi4right)tag3
$$
answered Mar 21 at 20:00
robjohn♦robjohn
270k27313641
$endgroup$
$begingroup$
Mathematica agrees with this answer.
$endgroup$
– robjohn♦
Mar 21 at 20:01
$begingroup$
Thank you sir, I have checked your answer, it is correct, just in the first sum is (4n)^2 and not 4n^2. Also my answer is also correct, i have checked it with wolfram mathematica also on yt youtube.com/watch?v=bG6bpYlQx2s
$endgroup$
– Davor
Mar 22 at 9:13
$begingroup$
@Davor: you mean that the first sum should be $sumlimits_n=1^inftyfrac116n^2-1$? Are you going to alter the question, or ask another with the correct sum?
$endgroup$
– robjohn♦
Mar 22 at 9:23
$begingroup$
Yes, i have corrected the equation, but it does not matter that much, i was more interested how to do it. We did not learn that approach with complex analysis, only usinf fourier transform
$endgroup$
– Davor
Mar 22 at 9:35
add a comment |
$begingroup$
The sum of the first term is
$$
beginalign
sum_n=1^inftyfrac1(4n)^2-1
&=frac12sum_n=1^inftyleft(frac14n-1-frac14n+1right)\
&=frac12left[1+sum_ninmathbbZfrac14n-1right]\
&=frac12+frac18sum_ninmathbbZfrac1n-frac14\
&=frac12-fracpi8cotleft(fracpi4right)\[6pt]
&=frac12-fracpi8tag1
endalign
$$
The sum of the second term is
$$
beginalign
sum_n=1^inftyfrac1(4n+2)^2+1
&=frac i2sum_n=1^inftyleft(frac1i-(4n+2)+frac1i+(4n+2)right)\
&=frac i2left[sum_ninmathbbZfrac1i+(4n+2)-frac1i+2-frac1i-2right]\
&=frac i8sum_ninmathbbZfrac1n+frac2+i4-frac15\
&=fracpi i8cotleft(pifrac2+i4right)-frac15\[3pt]
&=fracpi8tanhleft(fracpi4right)-frac15tag2
endalign
$$
where we have used $(7)$ from this answer (which uses complex analysis) or Theorem $2$ from this answer (which uses Fourier Series).
Subtracting $(2)$ from $(1)$ yields
$$
sum_n=1^inftyleft(frac1(4n)^2-1-frac1(4n+2)^2+1right)=frac710-fracpi8-fracpi8tanhleft(fracpi4right)tag3
$$
answered Mar 21 at 20:00
robjohn♦robjohn
270k27313641
$endgroup$
The sum of the first term is
$$
beginalign
sum_n=1^inftyfrac1(4n)^2-1
&=frac12sum_n=1^inftyleft(frac14n-1-frac14n+1right)\
&=frac12left[1+sum_ninmathbbZfrac14n-1right]\
&=frac12+frac18sum_ninmathbbZfrac1n-frac14\
&=frac12-fracpi8cotleft(fracpi4right)\[6pt]
&=frac12-fracpi8tag1
endalign
$$
The sum of the second term is
$$
beginalign
sum_n=1^inftyfrac1(4n+2)^2+1
&=frac i2sum_n=1^inftyleft(frac1i-(4n+2)+frac1i+(4n+2)right)\
&=frac i2left[sum_ninmathbbZfrac1i+(4n+2)-frac1i+2-frac1i-2right]\
&=frac i8sum_ninmathbbZfrac1n+frac2+i4-frac15\
&=fracpi i8cotleft(pifrac2+i4right)-frac15\[3pt]
&=fracpi8tanhleft(fracpi4right)-frac15tag2
endalign
$$
where we have used $(7)$ from this answer (which uses complex analysis) or Theorem $2$ from this answer (which uses Fourier Series).
Subtracting $(2)$ from $(1)$ yields
$$
sum_n=1^inftyleft(frac1(4n)^2-1-frac1(4n+2)^2+1right)=frac710-fracpi8-fracpi8tanhleft(fracpi4right)tag3
$$
answered Mar 21 at 20:00
robjohn♦robjohn
270k27313641
edited Mar 22 at 13:28
answered Mar 21 at 20:00
robjohn♦robjohn
270k27313641
answered Mar 21 at 20:00
robjohn♦robjohn
270k27313641
answered Mar 21 at 20:00
robjohn♦robjohn
270k27313641
270k27313641
$begingroup$
Mathematica agrees with this answer.
$endgroup$
– robjohn♦
Mar 21 at 20:01
$begingroup$
Thank you sir, I have checked your answer, it is correct, just in the first sum is (4n)^2 and not 4n^2. Also my answer is also correct, i have checked it with wolfram mathematica also on yt youtube.com/watch?v=bG6bpYlQx2s
$endgroup$
– Davor
Mar 22 at 9:13
$begingroup$
@Davor: you mean that the first sum should be $sumlimits_n=1^inftyfrac116n^2-1$? Are you going to alter the question, or ask another with the correct sum?
$endgroup$
– robjohn♦
Mar 22 at 9:23
$begingroup$
Yes, i have corrected the equation, but it does not matter that much, i was more interested how to do it. We did not learn that approach with complex analysis, only usinf fourier transform
$endgroup$
– Davor
Mar 22 at 9:35
add a comment |
$begingroup$
Mathematica agrees with this answer.
$endgroup$
– robjohn♦
Mar 21 at 20:01
$begingroup$
Thank you sir, I have checked your answer, it is correct, just in the first sum is (4n)^2 and not 4n^2. Also my answer is also correct, i have checked it with wolfram mathematica also on yt youtube.com/watch?v=bG6bpYlQx2s
$endgroup$
– Davor
Mar 22 at 9:13
$begingroup$
@Davor: you mean that the first sum should be $sumlimits_n=1^inftyfrac116n^2-1$? Are you going to alter the question, or ask another with the correct sum?
$endgroup$
– robjohn♦
Mar 22 at 9:23
$begingroup$
Yes, i have corrected the equation, but it does not matter that much, i was more interested how to do it. We did not learn that approach with complex analysis, only usinf fourier transform
$endgroup$
– Davor
Mar 22 at 9:35
$begingroup$
Mathematica agrees with this answer.
$endgroup$
– robjohn♦
Mar 21 at 20:01
$begingroup$
Mathematica agrees with this answer.
$endgroup$
– robjohn♦
Mar 21 at 20:01
$begingroup$
Thank you sir, I have checked your answer, it is correct, just in the first sum is (4n)^2 and not 4n^2. Also my answer is also correct, i have checked it with wolfram mathematica also on yt youtube.com/watch?v=bG6bpYlQx2s
$endgroup$
– Davor
Mar 22 at 9:13
$begingroup$
Thank you sir, I have checked your answer, it is correct, just in the first sum is (4n)^2 and not 4n^2. Also my answer is also correct, i have checked it with wolfram mathematica also on yt youtube.com/watch?v=bG6bpYlQx2s
$endgroup$
– Davor
Mar 22 at 9:13
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@Davor: you mean that the first sum should be $sumlimits_n=1^inftyfrac116n^2-1$? Are you going to alter the question, or ask another with the correct sum?
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– robjohn♦
Mar 22 at 9:23
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@Davor: you mean that the first sum should be $sumlimits_n=1^inftyfrac116n^2-1$? Are you going to alter the question, or ask another with the correct sum?
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– robjohn♦
Mar 22 at 9:23
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Yes, i have corrected the equation, but it does not matter that much, i was more interested how to do it. We did not learn that approach with complex analysis, only usinf fourier transform
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– Davor
Mar 22 at 9:35
$begingroup$
Yes, i have corrected the equation, but it does not matter that much, i was more interested how to do it. We did not learn that approach with complex analysis, only usinf fourier transform
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– Davor
Mar 22 at 9:35
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$begingroup$
$cos(n fracpi2) = 1 forall n in mathbbN$
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– Max
Mar 21 at 11:01
2
$begingroup$
@Max This is by factor 4 far from truth.
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– user
Mar 21 at 11:04
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Sorry, lack of coffee. Was meant to be $=0$.
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– Max
Mar 21 at 11:13
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Its not 0, it can be 0, -1 or 1 depending on n
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– Davor
Mar 21 at 11:20
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Oh dammit, i'm getting coffee now! Sorry for the nonsense.
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– Max
Mar 21 at 11:30